Difference between revisions of "2002 AMC 12B Problems/Problem 12"
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=== Solution 3 === | === Solution 3 === | ||
− | If <math>\frac{n}{20-n} = k^2 \ge 0</math>, then | + | If <math>\frac{n}{20-n} = k^2 \ge 0</math>, then <math>n \ge 0</math> and <math>20-n > 0</math>, otherwise <math>\frac{n}{20-n}</math> will be negative. Thus <math>0 \le n \le 19</math> and <cmath>0 = \frac{0}{20-(0)} \le \frac{n}{20-n} \le \frac{19}{20-(19)} = 19</cmath> Checking all <math>k</math> for which <math>0 \le k^2 \le 19</math>, we have 0, 1, 2, 3 as the possibilities. <math>\boxed{(D)}</math> |
~ Nafer | ~ Nafer |
Revision as of 14:42, 2 July 2019
- The following problem is from both the 2002 AMC 12B #12 and 2002 AMC 10B #16, so both problems redirect to this page.
Problem
For how many integers is the square of an integer?
Solution
Solution 1
Let , with (note that the solutions do not give any additional solutions for ). Then rewriting, . Since , it follows that divides . Listing the factors of , we find that are the only solutions (respectively yielding ).
Solution 2
For and the fraction is negative, for it is not defined, and for it is between 0 and 1.
Thus we only need to examine and .
For and we obviously get the squares and respectively.
For prime the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
This leaves , and a quick substitution shows that out of these only and yield a square. Therefore, there are only solutions (respectively yielding ).
Solution 3
If , then and , otherwise will be negative. Thus and Checking all for which , we have 0, 1, 2, 3 as the possibilities.
~ Nafer
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.